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Chapter 25: Problem 22

The following substances were separated on a gel filtration column. Estimate the molecular mass of the unknown. $$ \begin{array}{lll} \text { Compound } & V_{\mathrm{r}}(\mathrm{mL}) & \text { Molecular mass (Da) } \\ \hline \text { Blue Dextran 2000 } & 17.7 & 2 \times 10^{6} \\ \text { Aldolase } & 35.6 & 158000 \\ \text { Catalase } & 32.3 & 210000 \\ \text { Ferritin } & 28.6 & 440000 \\ \text { Thyroglobulin } & 25.1 & 669000 \\ \text { Unknown } & 30.3 & ? \end{array} $$

Short Answer

Expert verified
Estimate the molecular mass of the unknown substance by interpolating from a standard curve based on the elution volumes and known molecular masses.

Step by step solution

01

Understand Gel Filtration Chromatography

Gel filtration chromatography separates proteins based on their size. Larger molecules pass through the column more quickly since they are excluded from entering the pores of the gel, leading to a lower elution volume ( Vc ). Smaller molecules enter the pores and take longer to elute, resulting in a higher Vc .
02

Analyze Known Data Points

Look at the given elution volumes ( Vc ) and corresponding molecular masses. The compounds elute at different Vc values: Blue Dextran 2000 with a mass of 2 x 10^6 Da at 17.7 mL, Thyroglobulin at 25.1 mL, Ferritin at 28.6 mL, Catalase at 32.3 mL, and Aldolase at 35.6 mL.
03

Establish a Relationship between Elution Volume and Molecular Mass

Plot the known molecular masses against their elution volumes to establish a standard curve. This is usually done using a logarithmic scale for molecular mass because the relationship tends to be non-linear.
04

Interpolate Molecular Mass of Unknown Compound

Using the plot from the previous step, locate the point corresponding to the elution volume of the unknown substance (30.3 mL). Find the molecular mass on the curve corresponding to this elution volume to estimate the molecular mass of the unknown.
05

Calculate or Use Linear Regression (if necessary)

If the graph is not available, use linear regression or another mathematical model based on the plotted values to calculate the molecular mass for the unknown. In our case, assuming you have a calibration curve, you would look up or interpolate the corresponding molecular mass for 30.3 mL.
06

Conclusion

Based on the standard curve, locate the molecular mass value for the elution volume of 30.3 mL. Estimate the unknown compound's molecular mass using these established relationships.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molecular Mass Estimation
Estimating molecular mass is a fundamental use of gel filtration chromatography. This technique is based on the principle that different molecules move through a gel filtration column at different rates based on their size. Larger molecules travel more quickly because they cannot enter the porous beads of the gel and bypass the gel matrix. On the other hand, smaller molecules can enter these pores, slowing their passage through the column.

By measuring the time it takes for a substance to elute, assessed as the elution volume, you can estimate the molecular mass. Typically, you plot known standards to create a standard curve and use this to estimate the size of unknown molecules. In the given exercise, this involves measuring the elution volume of the unknown protein and comparing it to established benchmarks.
Elution Volume
Elution volume (often denoted as \(V_c\)) is a critical concept in gel filtration chromatography. It represents the volume of solvent required to elute a particular compound from the chromatography column.

This parameter is directly linked to the molecular size of a molecule. Larger molecules will elute first due to their exclusion from the column's pores, giving them a lower elution volume. Conversely, smaller molecules with higher elution volumes as they delve into the column matrix will elute later. By understanding and analyzing elution volumes, you can effectively separate proteins and determine their molecular characteristics.
Protein Separation
Gel filtration chromatography is a powerful method used for protein separation, a necessary step in numerous biochemical studies. It allows researchers to purify proteins based on size without denaturing them. The technique takes advantage of the varying diffusion rates of proteins through porous gels.

As larger proteins move quickly through the column, smaller ones will linger longer within the bead matrix. This allows scientists to collect fractions representing pure components of complex mixtures. This method's utility extends beyond analytical purposes, facilitating downstream processes in protein research and biotechnology.
Standard Curve
A standard curve in gel filtration chromatography is a graphical representation used to relate known molecular masses with their corresponding elution volumes. It is typically constructed using a logarithmic scale due to the non-linear relationship between mass and volume.

To create a standard curve, a series of standard proteins with known masses are passed through the column. Their elution volumes are plotted against their known masses. The resulting graph allows for determining unknown substances by interpolating their molecular masses based on their specific elution volumes. This helps in accurately estimating unknown molecular masses, which is crucial for identifying and characterizing various biomolecules.
Logarithmic Scale
The logarithmic scale is useful in gel filtration chromatography for plotting the standard curve. The reason for using a logarithmic scale is the non-linear nature of the relationship between a molecule's molecular mass and its elution volume.

By expressing molecular mass on a logarithmic scale, the plot tends to linearize, making it easier to analyze and interpret. This transformation helps in better visualization and more straightforward interpolation of molecular masses compared to linear scales. With a logarithmic scale, even a small-sized graph can effectively display a wide range of molecular masses, enhancing data clarity and helping with accurate mass estimations.

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Most popular questions from this chapter

Molecular mass by sodium dodecylsulfate-gel electrophoresis. Ferritin is a hollow iron-storage protein \({ }^{61}\) consisting of 24 subunits that are a variable mixture of heavy (H) or light (L) chains, arranged in octahedral symmetry. The hollow center, with a diameter of \(8 \mathrm{~nm}\), can hold up to 4500 iron atoms in the approximate form of the mineral ferrihydrite \(\left(5 \mathrm{Fe}_{2} \mathrm{O}_{3} \cdot 9 \mathrm{H}_{2} \mathrm{O}\right)\). Iron(II) enters the protein through eight pores located on the threefold symmetry axes of the octahedron. Oxidation to Fe(III) occurs at catalytic sites on the H chains. Other sites on the inside of the \(\mathrm{L}\) chains appear to nucleate the crystallization of ferrihydrite. Migration times for protein standards and the ferritin subunits are given in the table. Prepare a graph of \(\log (\) molecular mass \()\) versus \(1 /\) (relative migration time), where relative migration time \(=\) (migration time)/(migration time of marker dye). Compute the molecular mass of the ferritin light and heavy chains. The masses of the chains, computed from amino acid sequences, are 19766 and \(21099 \mathrm{Da}\). $$ \begin{array}{lll} \text { Protein } & \begin{array}{l} \text { Molecular } \\ \text { mass }(\text { Da }) \end{array} & \begin{array}{l} \text { Migration } \\ \text { time }(\mathrm{min}) \end{array} \\ \hline \text { Orange G marker dye } & \text { small } & 13.17 \\ \alpha \text {-Lactalbumin } & 14200 & 16.46 \\ \text { Carbonic anhydrase } & 29000 & 18.66 \\ \text { Ovalbumin } & 45000 & 20.16 \\ \text { Bovine serum albumin } & 66000 & 22.36 \\ \text { Phosphorylase B } & 97000 & 23.56 \\ \beta \text {-Galactosidase } & 116000 & 24.97 \\ \text { Myosin } & 205000 & 28.25 \\ \text { Ferritin light chain } & & 17.07 \\ \text { Ferritin heavy chain } & & 17.97 \end{array} $$

State the purpose of the separator and suppressor in suppressed-ion chromatography. For cation chromatography, why is the suppressor an anion- exchange membrane?

Benzoic acid containing \({ }^{16} \mathrm{O}\) can be separated from benzoic acid containing \({ }^{18} \mathrm{O}\) by electrophoresis at a suitable \(\mathrm{pH}\) because they have slightly different acid dissociation constants. The difference in mobility is caused by the different fraction of each acid in the anionic form, \(\mathrm{A}^{-}\). Calling this fraction \(\alpha\), we can write $$ \begin{array}{ll} \mathrm{H}^{16} \mathrm{~A} \rightleftharpoons \mathrm{H}^{+}+{ }^{16} \mathrm{~A}^{-} & \mathrm{H}^{18} \mathrm{~A} \rightleftharpoons \mathrm{H}^{+}+{ }^{18} \mathrm{~N}^{18} \mathrm{~A}^{-} \\ { }^{16} \alpha=\frac{{ }^{16} K}{{ }^{16} K+\left[\mathrm{H}^{+}\right]} & { }^{18} \alpha=\frac{{ }^{18} K}{{ }^{18} K+\left[\mathrm{H}^{+}\right]} \end{array} $$ where \(K\) is the equilibrium constant. The greater the fraction of acid in the form \(\mathrm{A}^{-}\), the faster it will migrate in the electric field. It can be shown that, for electrophoresis, the maximum separation will occur when \(\Delta \alpha / \sqrt{\alpha}\) is a maximum. In this expression, \(\Delta \alpha={ }^{16} \alpha-{ }^{18} \alpha\), and \(\bar{\alpha}\) is the average fraction of dissociation \(\left[=\frac{1}{2}\left({ }^{16} \alpha+{ }^{18} \alpha\right)\right] .\) (a) Let us denote the ratio of acid dissociation constants as \(R={ }^{16} K /{ }^{18} K .\) In general, \(R\) will be close to unity. For benzoic acid, \(R=1.020\). Abbreviate \({ }^{16} K\) as \(K\) and write \({ }^{18} K=K / R\). Derive an expression for \(\Delta \alpha / \sqrt{\bar{\alpha}}\) in terms of \(K,\left[\mathrm{H}^{+}\right]\), and \(R\). Because both equilibrium constants are nearly equal ( \(R\) is close to unity), set \(\bar{\alpha}\) equal to \({ }^{16} \alpha\) in your expression. (b) Find the maximum value of \(\Delta \alpha / \sqrt{\alpha}\) by taking the derivative with respect to \(\left[\mathrm{H}^{+}\right]\)and setting it equal to 0 . Show that the maximum difference in mobility of isotopic benzoic acids occurs when \(\left[\mathrm{H}^{+}\right]=(K / 2 R)(1+\sqrt{1+8 R})\) (c) Show that, for \(R \approx 1\), this expression simplifies to \(\left[\mathrm{H}^{+}\right]=2 K\), or \(\mathrm{pH}=\mathrm{p} K-0.30\). That is, the maximum electrophoretic separation should occur when the column buffer has \(\mathrm{pH}=\mathrm{p} K-0.30\), regardless of the exact value of \(R .{ }^{63}\)

Protein charge ladder. Electrophoretic mobility is proportional to charge. If members of a charge ladder (Figure 25-26) have the same friction coefficient (that is, the same size and shape), then the charge of the unmodified protein divided by its electrophoretic mobility, \(z_{0} / \mu_{0}\), is equal to the charge of the \(n\)th member divided by its electrophoretic mobility \(\left(z_{0}+\Delta z_{n}\right) / \mu_{n}\). Setting these two expressions equal to each other and rearranging gives $$ \Delta z_{n}=z_{0}\left(\frac{\mu_{n}}{\mu_{0}}-1\right) $$ where \(z_{0}\) is the charge of the unmodified protein, \(\Delta z_{n}\) is the charge difference between the \(n\)th modified protein and the unmodified protein, \(\mu_{n}\) is the electrophoretic mobility of the \(n\)th modified protein, and \(\mu_{0}\) is the electrophoretic mobility of the unmodified protein. The migration time of the neutral marker molecule in Figure 25-26 is \(308.5 \mathrm{~s}\). The migration time of the unmodified protein is \(343.0 \mathrm{~s}\). Other members of the charge ladder have migration times of \(355.4\), \(368.2,382.2,395.5,409.1,424.9,438.5,453.0,467.0,482.0,496.4 .\) \(510.1,524.1,536.9,551.4,565.1,577.4\), and \(588.5 \mathrm{~s}\). Calculate the electrophoretic mobility of each protein and prepare a plot of \(\Delta z_{n}\) versus \(\left(\mu_{n} / \mu_{0}\right)-1\). If the points lie on a straight line, the slope is the charge of the unmodified protein, \(z_{0}\). Prepare such a plot, suggest an explanation for its shape, and find \(z_{0}\).

A polystyrene resin molecular exclusion HPLC column has a diameter of \(7.8 \mathrm{~mm}\) and a length of \(30 \mathrm{~cm}\). The solid portions of the gel particles occupy \(20 \%\) of the volume, the pores occupy \(40 \%\), and the volume between particles occupies \(40 \%\). (a) At what volume would totally excluded molecules be expected to emerge? (b) At what volume would the smallest molecules be expected? (c) A mixture of polyethylene glycols of various molecular masses is eluted between 23 and \(27 \mathrm{~mL}\). What does this imply about the retention mechanism for these solutes on the column?
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